The given function is tan 3 2xsec2x.
Given equation can be rewritten as,
tan 3 2xsec2x= tan 2 2xtan2xsec2x =( sec 2 2x−1)tan2xsec2x = sec 2 2xtan2xsec2x−tan2xsec2x
∫ ( tan 3 2xsec2x )dx = ∫ ( sec 2 2xtan2xsec2x−tan2xsec2x ) dx = ∫ sec 2 2xtan2xsec2xdx− ∫ tan2xsec2xdx = ∫ sec 2 2xtan2xsec2xdx− sec2x 2 +c (1)
Let, sec2x=t
Differentiate with respect to x,
2sec2xtan2xdx=dt
Substitute t in (1),
∫ ( tan 3 2xsec2x )dx = ∫ 1 2 t 2 − sec2x 2 +c = t 3 6 − sec(2x) 2 +c = (sec(2x)) 3 6 − sec(2x) 2 +c
Thus, the integral of the function tan 3 2xsec2x is (sec(2x)) 3 6 − sec(2x) 2 +c.